SECTION 11.3 Solving Quadratic Equations b Graphing 11.3 OBJECTIVES 1. Find an ais of smmetr 2. Find a verte 3. Graph a parabola 4. Solve quadratic equations b graphing 5. Solve an application involving a quadratic equation In Section 3.3, we learned to graph a linear equation. We discovered that the graph of ever linear equation was a straight line. In this section, we will consider the graph of a quadratic equation. Consider an equation of the form a 2 b c a 0 This equation is quadratic in and linear in. Its graph will alwas be the curve called the parabola. In an equation of the form a 2 b c a 0 the parabola opens upward or downward, as follows: 1. If a 0, the parabola opens upward. 2. If a 0, the parabola opens downward. a 2 b c Verte a > 0 a < 0 Verte Two concepts regarding the parabola can be made b observation. Consider the above illustrations. 741
742 Chapter 11 Quadratic Functions 1. There is alwas a minimum (or lowest) point on the parabola if it opens upward. There is alwas a maimum (or highest) point on the parabola if it opens downward. In either case, that maimum or minimum value occurs at the verte of the parabola. 2. Ever parabola has an ais of smmetr. In the case of parabolas that open upward or downward, that ais of smmetr is a vertical line midwa between an pair of smmetric points on the parabola. Also, the point where this ais of smmetr intersects the parabola is the verte of the parabola. The following figure summarizes these observations. Ais of smmetr Equal distances Smmetric points Verte Our objective is to be able to quickl sketch a parabola. This can be done with as few as three points if those points are carefull chosen. For this purpose ou will want to find the verte and two smmetric points. First, let s see how the coordinates of the verte can be determined from the standard equation a 2 b c (1) In equation (1), if 0, then c, and so (0, c) gives the point where the parabola intersects the ais (the intercept). Look at the sketch. To determine the coordinates of the smmetric point ( 1, c), note that it lies along the horizontal line c. Smmetric points c (0, c) ( 1, c) Verte
Section 11.3 Solving Quadratic Equations b Graphing 743 Therefore, let c in equation (1): c a 2 b c 0 a 2 b 0 (a b) and 0 or b a We now know that (0, c) and b a, c are the coordinates of the smmetric points shown. Since the ais of smmetr must be midwa between these points, the value along that ais is given b 0 ( b/a) b (2) 2 2 a Since the verte for an parabola lies on the ais of smmetr, we can now state the following general result. Verte of a Parabola If Note: The coordinate of the verte can be found most easil b substituting the value found for into the original equation. a 2 b c a 0 then the coordinate of the verte of the corresponding parabola is b 2 a We now know how to find the verte of a parabola, and if two smmetric points can be determined, we are well on our wa to the desired graph. Perhaps the simplest case is when the quadratic member of the given equation is factorable. In most cases, the two intercepts will then give two smmetric points that are ver easil found. Eample 1 illustrates such a case.
744 Chapter 11 Quadratic Functions E a m p l e 1 Sketch the information to help ou solve the problem. Begin b drawing as a dashed line the ais of smmetr. Graphing a Parabola Graph the equation 2 2 8 First, find the ais of smmetr. In this equation, a 1, b 2, and c 8. We then have 1 b 2 2 a 2 1 2 2 1 Thus, 1 is the ais of smmetr. Second, find the verte. Since the verte of the parabola lies on the ais of smmetr, let 1 in the original equation. If 1, ( 1) 2 2( 1) 8 9 At this point ou can plot the verte along the ais of smmetr. and ( 1, 9) is the verte of the parabola. Third, find two smmetric points. Note that the quadratic member in this case is factorable, and so setting 0 in the original equation will quickl give two smmetric points (the intercepts): 0 2 2 8 ( 4)( 2) So when 0, ( 1, 9) ( 4, 0) (2, 0) ( 1, 9) 4 0 or 2 0 4 2 and our intercepts are ( 4, 0) and (2, 0). Fourth, draw a smooth curve connecting the points found above, to form the parabola. Note: You can choose to find additional pairs of smmetric points at this time if necessar. For instance, the smmetric points (0, 8) and ( 2, 8) are easil located. CHECK YOURSELF 1 Graph the equation 2 2 3 (Hint: Since the coefficient of 2 is negative, the parabola opens downward.)
Section 11.3 Solving Quadratic Equations b Graphing 745 A similar process will work if the quadratic member of the given equation is not factorable. In that case, one of two things happens: 1. The intercepts are irrational and therefore not particularl helpful in the graphing process. 2. The intercepts do not eist. Consider Eample 2. E a m p l e 2 Graphing a Parabola Graph the function f() = 3 f() 2 6 3 First, find the ais of smmetr. Here a 1, b 6, and c 3. So Thus, 3 is the ais of smmetr. Second, find the verte. If 3, b ( 6) 6 2 a 2 1 2 3 f(3) 3 2 6 3 3 6 (3, 6) and (3, 6) is the verte of the desired parabola. Third, find two smmetric points. Here the quadratic member is not factorable, so we need to find another pair of smmetric points. f() Smmetric points (0, 3) (6, 3) intercept 3 units 3 units Note that (0, 3) is the intercept of the parabola. We found the ais of smmetr at 3 in step 1. Note that the smmetric point to (0, 3) lies along the horizontal line through the intercept at the same distance (3 units) from the ais of smmetr. Hence, (6, 3) is our smmetric point. Fourth, draw a smooth curve connecting the points found above to form the parabola.
746 Chapter 11 Quadratic Functions f() (0, 3) (6, 3) Note: An alternate method is available in step 3. Observing that 3 is the intercept and that the smmetric point lies along the line 3, set f() 3 in the original equation: 3 2 6 3 0 2 6 0 ( 6) so (3, 6) 0 or 6 0 6 and (0, 3) and (6, 3) are the desired smmetric points. CHECK YOURSELF 2 Graph the function f() 2 4 5 Thus far the coefficient of 2 has been 1 or 1. The following eample shows the effect of different coefficients on the shape of the graph of a quadratic equation. E a m p l e 3 Graphing a Parabola Graph the equation 3 2 6 5 First, find the ais of smmetr. (0, 5) (2, 5) b ( 6) 6 2 a 2 3 6 1 Second, find the verte. If 1, (1, 2) 3(1) 2 6 1 5 2 So (1, 2) is the verte. Third, find smmetric points. Again the quadratic member is not factorable, and we use the intercept (0, 5) and its smmetric point (2, 5).
Fourth, connect the points with a smooth curve to form the parabola. Compare this curve to those in previous eamples. Note that the parabola is tighter about the ais of smmetr. That is the effect of the larger 2 coefficient. CHECK YOURSELF 3 Graph the equation Section 11.3 Solving Quadratic Equations b Graphing 747 1 2 2 3 1 The following algorithm summarizes our work thus far in this section. To Graph a Parabola Step 1 Step 2 Step 3 Step 4 Find the ais of smmetr. Find the verte. Determine two smmetric points. Note: You can use the intercepts if the quadratic member of the given equation is factorable. Otherwise use the intercept and its smmetric point. Draw a smooth curve connecting the points found above to form the parabola. You ma choose to find additional pairs of smmetric points at this time. We have seen that quadratic equations can be solved in three different was: b factoring (Section 6.5), b completing the square (Section 11.1), or b using the quadratic formula (Section 11.2). We will now look at a fourth technique for solving quadratic equations, a graphical method. Unlike the other methods, the graphical technique ma ield onl an approimation of the solution(s). On the other hand, it can be a ver useful method for checking the reasonableness of eact answers. This is particularl true when technolog is used to produce the graph. We can easil use a graph to identif the number of positive and negative solutions to a quadratic equation. Recall that solutions to the equation 0 a 2 b c are values for that make the statement true. If we have the graph of the parabola a 2 b c then values for which is equal to zero are solutions to the original equation. We know that 0 is the equation for the ais. Solutions for the equation eist wherever the graph crosses the ais.
748 Chapter 11 Quadratic Functions E a m p l e 4 Identifing the Number of Solutions to a Quadratic Equation Find the number of positive solutions and the number of negative solutions for each quadratic equation. (a) 0 2 3 7 Using either the methods of this section, or a graphing device, we find the graph of the parabola 2 3 7 first. The ais of smmetr is 3. Two points on the graph could be (0, 7) and ( 3, 7). 2 Looking at the graph, we see that the parabola crosses the ais twice, once to the left of zero and once to the right. There are two real solutions. One is negative and one is positive. Note that due to the basic shape of a parabola, it is not possible for this graph to cross the ais more than two times. (b) 0 5 2 32 Again, we graph the parabola 5 2 32 first. The line of smmetr is 1 6. 2 5 Two points on the parabola are (0, 0) and 3, 0 5. 400 300 200 100 10 5 5 10 There are two solutions. One is positive and the other appears to be zero. A quick check of the equation confirms that zero is a solution.
Section 11.3 Solving Quadratic Equations b Graphing 749 (c) 0 2 2 3 The ais of smmetr is 1. Two points on the parabola would be (0, 3) and ( 2, 3). Here is the graph of the related parabola. Notice that the graph never touches the ais. There are no real solutions to the equation 0 2 2 3. CHECK YOURSELF 4 Use the graph of the related parabola to find the number of real solutions to each equation. Identif the number of solutions that are positive and the number that are negative. (a) 0 3 2 16 (b) 0 2 4 (c) 0 2 3 9 We used the intercepts of the graph to determine the number of solutions for each equation in Eample 4. The value of the coordinate for each intercept is the eact solution. In the net eample, we ll use the graphical method to estimate the solutions of a quadratic equation. E a m p l e 5 Solving a Quadratic Equation Solve each equation. (a) 0 2 3 7
750 Chapter 11 Quadratic Functions As we did in Eample 4, we find the graph of the parabola 2 3 7 first. In Eample 4, we pointed out that there are two solutions. One solution is between 5 and 4. The other is between 1 and 2. The precision with which we estimate the values depends on the qualit of the graph we are eamining. From this graph, 4.5 and 1.5 would be reasonable estimates of the solutions. Using the quadratic formula, we find that the actual solutions are ( 3 3 7 ) 2 and ( 3 3 7 ). Using a calculator, we can round these answers to the nearest hundredth. We get 1.54 and 4.54. Our estimates were quite 2 reasonable. (b) 0 5 2 32 Again, we graph the parabola 5 2 32 first. 400 300 200 100 10 5 5 10 There are two solutions. One is a little more than si and the other appears to be zero. Using the quadratic formula (or factoring) we find that the solutions to this equation are 0 and 6.4.
Section 11.3 Solving Quadratic Equations b Graphing 751 CHECK YOURSELF 5 Use the graph of the related parabola to estimate the solutions to each equation. (a) 0 3 2 16 (b) 0 2 3 9 From graphs of equations of the form a 2 b c, we know that if a 0, then the verte is the lowest point on the graph (the minimum value). Also, if a 0, then the verte is the highest point on the graph (the maimum value). We can use this result to solve a variet of problems in which we want to find the maimum or minimum value of a variable. The following are just two of man tpical eamples. E a m p l e 6 An Application Involving a Quadratic Function A software compan sells a word processing program for personal computers. The have found that their monthl profit in dollars, P, from selling copies of the program is approimated b P() 0.3 2 90 1500 Find the number of copies of the program that should be sold in order to maimize the profit. Since the relating equation is quadratic, the graph must be a parabola. Also since the coefficient of 2 is negative, the parabola must open downward, and thus the verte will give the maimum value for the profit, P. To find the verte, b 90 90 150 2 a 2( 0.3) 0. 6 The maimum profit must then occur when 150, and we substitute that value into the original equation: P() 0.3(150) 2 (90)(150) 1500 $5250 The maimum profit will occur when 150 copies are sold per month, and that profit will be $5250.
752 Chapter 11 Quadratic Functions CHECK YOURSELF 6 A compan that sells portable radios finds that its weekl profit in dollars, P, and the number of radios sold,, are related b P() 0.2 2 40 100 Find the number of radios that should be sold to have the largest weekl profit. Also find the amount of that profit. E a m p l e 7 An Application Involving a Quadratic Function A farmer has 3600 ft of fence and wishes to enclose the largest possible rectangular area with that fencing. Find the largest possible area that can be enclosed. As usual, when dealing with geometric figures, we start b drawing a sketch of the problem. Length Width First, we can write the area, A, as Area length width A (3) Also, since 3600 ft of fence is to be used, we know that The perimeter of the region is 2 2 2 2 3600 2 3600 2 (4) 1800 Substituting for in equation (3), we have A() (1800 ) 1800 2 2 1800 (5)
Section 11.3 Solving Quadratic Equations b Graphing 753 The width is 900 ft. Since from equation (2) 1800 900 900 ft The length is also 900 ft. The desired region is a square. Again, the graph for A is a parabola opening downward, and the largest possible area will occur at the verte. As before, and the largest possible area is 1800 1800 900 2( 1) 2 A() (900) 2 1800(900) 810,000 ft 2 CHECK YOURSELF 7 We want to enclose three sides of the largest possible rectangular area b using 900 ft of fence. Assume that an eisting wall makes the fourth side. What will be the dimensions of the rectangle? CHECK YOURSELF ANSWERS 1. 2. ( 1, 4) ( 4, 5) (0, 5) ( 3, 0) (1, 0) ( 2, 1) 3. (0, 1) (6, 1) (3, 5.5) 4. (a) Two solutions; one positive, one zero; (b) no real solutions; (c) two solutions; one negative, one positive. 5. (a) 0, 5.3; (b) 1.9, 4.9. 6. 100 radios, $1900. 7. Width 225 ft, length 450 ft.
754 Chapter 11 Quadratic Functions Electric Power Costs. You and a partner in a small engineering firm have been asked to help calculate the costs involved in running electric power a distance of 3 miles from a public utilit compan to a nearb communit. You have decided to consider three tower heights: 75 feet, 100 feet, and 120 feet. The 75-ft towers cost $850 each to build and install; the 100-ft towers cost $1110 each; and the 120-ft towers cost $1305 each. Spans between towers tpicall run 300 to 1200 ft. The conducting wires weigh 1900 pounds per 1000 ft, and the tension on the wires is 6000 pounds. To be safe, the conducting wires should never be closer than 45 ft to the ground. Work with a partner to complete the following. 1. Develop a plan for the communit showing all three scenarios and their costs. Remember that hot weather will epand the wires and cause them to sag more than normall. Therefore, allow for about a 10% margin of error when ou calculate the sag amount. 2. Write an accompaning letter to the town council eplaining our recommendation. Be sure to include an equations ou used to help ou make our decision. Remember to look at the formula at the beginning of the chapter, which gives the amount of sag for wires given a certain weight per foot, span length, and tension on the conductors.
E e r c i s e s 11.3 In Eercises 1 to 8, match each graph with one of the equations at the left. (a) 2 2 (b) 2 2 1 (c) 2 1 (d) 2 3 (e) 2 4 (f) 2 1 (g) 2 2 3 (h) 2 6 8 1. 2. 3. 4. 5. 6. 7. 8. 755
756 Chapter 11 Quadratic Functions In Eercises 9 to 14, which of the given conditions appl to the graphs of the following equations? Note that more than one condition ma appl. (a) The parabola opens upward. (b) The parabola opens downward. (c) The parabola has two intercepts. (d) The parabola has one intercept. (e) The parabola has no intercept. 9. 2 3 10. 2 4 11. 2 3 4 12. 2 2 2 13. 2 3 10 14. 2 8 16 In Eercises 15 to 28, find the equation of the ais of smmetr, the coordinates of the verte, and the intercepts. Sketch the graph of each equation. 15. 2 4 16. 2 1 17. 2 4 18. 2 2 19. 2 2 20. 2 3 21. 2 6 5 22. 2 6 23. 2 5 6 24. 2 6 5 25. 2 6 8 26. 2 3 4 27. 2 6 5 28. 2 6 8
Section 11.3 Solving Quadratic Equations b Graphing 757 In Eercises 29 to 40, find the equation of the ais of smmetr, the coordinates of the verte, and at least two smmetric points. Sketch the graph of each equation. 29. 2 2 1 30. 2 4 6 31. 2 4 1 32. 2 6 5 33. 2 3 3 34. 2 5 3 35. 2 2 4 1 36. 1 2 2 1 37. 1 3 2 3 38. 2 2 4 1 39. 3 2 12 5 40. 3 2 6 1 In Eercises 41 to 50, use the graph of the related parabola to find the number of real solutions to each equation. Identif the number of solutions that are positive and the number that are negative. 41. 0 2 12 42. 0 2 3 2 43. 0 6 2 19 44. 0 7 2 15 45. 0 2 2 5 46. 0 2 6 4 47. 0 9 2 12 7 48. 0 3 2 9 5 49. 0 2 2 7 50. 0 2 8 11 In Eercises 51 to 58, use the graph of the related parabola to estimate the solutions to each equation. Round answers to the nearest tenth. 51. 0 2 12 52. 0 2 3 2 53. 0 6 2 19 54. 0 7 2 15 55. 0 9 2 12 7 56. 0 3 2 9 5 57. 0 2 2 7 58. 0 2 8 11
758 Chapter 11 Quadratic Functions 59. Profit. A compan s weekl profit, P, is related to the number of items sold b P() 0.3 2 60 400. Find the number of items that should be sold each week in order to maimize the profit. Then find the amount of that weekl profit. 60. Profit. A compan s monthl profit, P, is related to the number of items sold b P() 0.2 2 50 800. How man items should be sold each month to obtain the largest possible profit? What is the amount of that profit? 61. Area. A builder wants to enclose the largest possible rectangular area with 2000 ft of fencing. What should be the dimensions of the rectangle, and what will the area of the rectangle be? 62. Area. A farmer wants to enclose a rectangular area along a river on three sides. If 1600 ft of fencing is to be used, what dimensions will give the maimum enclosed area? Find that maimum area. 63. Motion. A ball is thrown upward into the air with an initial velocit of 96 ft/s. If h gives the height of the ball at time t, then the equation relating h and t is h(t) 16t 2 96t Find the maimum height the ball will attain. 64. Motion. A ball is thrown upward into the air with an initial velocit of 64 ft/s. If h gives the height of the ball at time t, then the equation relating h and t is h(t) 16t 2 64t Find the maimum height the ball will attain. 65. Motion. A ball is thrown upward with an initial velocit v. After 3 s, it attains a height of 120 ft. Find the initial velocit using the equation h(t) 16t 2 v t For each of the following quadratic functions, use our graphing calculator to determine (a) the verte of the parabola and (b) the range of the function. 66. f() 2( 3) 2 1 67. g() 3( 4) 2 2 68. f() ( 1) 2 2 69. g() ( 2) 2 1 70. f() 3( 1) 2 2 71. g() 2( 4) 2 72. Eplain how to determine the domain and range of the function f() a( h) 2 k. In Eercises 73 to 76, describe a viewing window that would include the verte and all intercepts for the graph of each function. 73. f() 3 2 25 74. f() 9 2 5 7 75. f() 2 2 5 7 76. f() 5 2 2 7